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Dirction of fastest growth

  1. Nov 15, 2005 #1
    I have f(x,y,z)=(x^2)y-x(e^z) and point Po=(2,-1,pi)
    I need to find
    a) gradient at point Po ( done)
    b) Rate of change of f at point Po in the direction of vector u=i-2j+k (it's also done)
    c) Unit vector in the direction of fastest growth of f at Po.

    I can't find formulas for a last on. Does it come from a) and b)?
    I know that angle should be zero but I am not sure what angle it is.
  2. jcsd
  3. Nov 15, 2005 #2
    At what angle is the growth greatest? Its got to do with the directional derivative and the gradient.
  4. Nov 15, 2005 #3
    The growth is greatest when angle equals zero.
  5. Nov 15, 2005 #4


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    When WHAT angle equals zero?

    (You are almost there!)
  6. Nov 15, 2005 #5
    I think it's the angle between unit vector and tangent vector to f at Po.
    If I'm correct, I'm not sure how should I put it on a paper.
  7. Nov 15, 2005 #6
    Youre right. When you take a directional derivative what kind of product do you use (for the two vectors). One of these two is at a maximum when the angle is zero, thats the one you want to use.

    If you're still stuck, show the defining expression for a directional derivative.
    Last edited: Nov 15, 2005
  8. Nov 15, 2005 #7


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    Not "tangent vector" because you are not talking about a curve.

    The gradient is the "derivative" of a function of 2 or more variables. The derivative of f(x,y) in the direction of angle [itex]\theta[/itex] if given by
    [tex]D_\theta f(x,y)= (cos \theta i+ sin \theta j)\dot(\frac{\partial f}{\partial x}i+ \frac{\partial f}{\partial y})[/tex]
    That will have a maximum (with respect to [itex]\theta[/itex]) where it's derivative with respect to [itex]\theta[/itex] equals 0:
    [tex]-cos\theta \frac{\partial f}{\partial x}+ sin\theta\frac{\partial f}{\partial y}= 0[/tex]
    That means that
    [tex]tan\theta= \frac{sin\theta}{cos\theta}= \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}[/tex]

    Think about what that means in terms of the components of the gradient of f.
  9. Nov 15, 2005 #8
    It is tangent to the surface at that point and I think that is what he is meaning, but when youre dotting [itex] D_{\theta}f(x,y) [/itex] why did we get a negative cosine?

    You also forgot your j unit vector in the first expression.
  10. Nov 15, 2005 #9
    Yes, my mistake I thought of tangent vector to a surface.
    Why should I dot f(x,y) if I have f(x,y,z)?
    Still confused.
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